force 2.10-2.12

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Form 4© COPYRIGHT BY SUPER EDUCATION GROUP (JO

2.10 – 2.12

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F

% %

F&'

F

'

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UNDERSTANDING WORK, ENERGY AND EFFICIENCY

A Work

1. Work is the product of applied force and dis!"#ement in the direction of the applied force.

2. When the work is done ener$y is transferred from one object to another.

3. The work done is equal to the amount of ener$y transferred.

4. The SI unit for work is %o&!e.

The formulae of work

WORK = FORCE X DISPLACEMENT

W = F ! s

No work is done w'en(

)* The object is stationary

2) The direction of motion of the object is perpendicular to that of the applied force


W : work in Joule/J

F : force in Newton/N

s : displacement in

Work done

Force and displacement in the samedirection

Force and displacement in different directions

W = F.s W = Work F = Force s = displacement

W = FX . sW = F s cos θ W = work F = force s = displacement

θ = angle between forceand displacement

F

F Y F Y

F

F&

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Diagram (a) Diagram (b)

Diagrams (a) and (b) shows a boy pshing a load and a weightlifter lifting aload of !" kg

a) #alclate the work done

i. by the boy

W = F.s = 20 x 2 = 40 Nm or 40

ii. by the weightlifter

W = F.s = m!h

= "0 x #0 x 2 = #200 Nm or #200

$. %&man is plling a bo' with a force of " at an angle of !"o from the hori&ontal.#alclate the work done to mo*e the bo' to a distance of + m.

Wor$ = %omponent of force x displacement&'n the direction of displacement)

= (0 cos "0o x = 2( x = *(

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"!ample 1

"!ample 2

#$ %

&$$

s ' ( m

"!ample 3

"!ample 4

) ' &$$ %

S ' $.# m


W = Fs

If, F = 40 N and s = 2 m

Hence, W = 40 2

= !0 J

For+$ F

s

W = Fs

= !0 cos "00 #$%

= !0 #0&$% #$%

= 200 J

' '

F = (0 N

) = *&$ m

W = F s = F )

= (0 #*&$%

= 4$&0 J

W = F s

= "00 0&!

= 4!0 J

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For+$ F

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Ener$y

1. "ner*+ is the potential to do work.

2. "ner*+ cannot be created nor be destro+ed.

3. "!ist in ,arious forms such as otenti"! ener$y, kineti# ener$y- electrical ener*+- sound ener*+-

nuclear ener*+- heat and chemical ener*+

"ner*+ is de+ined as the capacit+ to do work.

Work is done when ener$y is #onerted from one form to another.

The unit of work is Nm or -o&!e.-*

A/Work done "nd t'e #'"n$e in kineti# ener$y

1. inetic ener*+ is ener*+ of an object due to its motion&

2. /efer to the fi*ure abo,e-

3. "!ample 0 small car of mass 1$$ k* is mo,in* alon* a flat road. The

resultant force on the car is 2$$ %.

a What is its kinetic ener*+ of the car after mo,in* throu*h 1$ m b What is its ,elocit+ after mo,in* throu*h 1$ m

+olution : i-en : m = *00 k. , F = 200 N

a& inetic ener., 1k = Fs

= 200 *0= 2000 J

& 3elocit, - 5 m- 2 = 2000

- = "&(2 m s

6*


Work ' )s

' mas

' m ,2

The formulae of inetic ener*+- "k ' m,2

Through, v 2 = u2 +2as

u = 0

and, as = ½ v 2

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0/ Work done "nd $r"it"tion"! otenti"! ener$y

h ' 1.( m

1. 5ra,itational potential ener*+ is ener*+ of an object due to its position. possessed b+ an object due to

its position in a *ra,itational field

2. /efer to the fi*ure abo,e

W ' F s ' m$ h

where- F 1 m$

So, Gr"it"tion"! ener$y, E 1 m$'


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2rin#i!e o+ #onser"tion o+ ener$y

the principle of conser,ation of ener*+.

1. "ner*+ cannot be created or destro+ed but can be chan*ed from one form to another form.

2. E3"m!e 6 a thrown ball upwards will achie,e a ma!imum hei*ht before chan*in* its direction and falls

3. "!ample in calculation 6 0 coconut falls from a tree from a hei*ht of 2$ m. What is the ,elocit+ of

coconut just before hittin* the earthi-en : ) = 20 m, u = 0 , . = 7&! ms62 , - = 8

1 p = 1k

m.) = 5 m- 2

m#7&!%#20% = 5m- 2

- 2 = (72, - = *7&! m s6*

2ower

1. 7ower is the rate of doing or!"

rate of energ# transfor$ation"

')erefore, power, 9 =ti$eta!en

or!done

, so, 9 =t

W

W)ere, 9 : power in watt/W

W : work in oule/J

t : time to do work in seconds/s

2. 8nit6 9oule per second 9 s:1 or Watt W

3. 0 wei*htlifter lifts 1#$ k* of wei*hts from the floor to a hei*ht of 2 m abo,e his head in a time of $.# s.

What is the ower $ener"ted b+ the wei*htlifter durin* this time

* ' ;.# ms:2


;aimum 9otential ener.

inetic ener. decrease potential ener. decrease

and potential ener. and kinetic ener.

Increase increase

;aimum

kinetic ener.

+olution : i-en : m = *!0 k., ) = 2 m, t = 0&! s and . = 7&! ms62& 9 = 8

9 =t

W

=t

$gh

=

0.8

2 9.8 180 ××

= 4 4*0 W

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E++i#ien#y

1.

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2rin#i!e ++ #onser"tion o+ ener$y


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,amy releases a $ kg metal ball from a bilding -" m high (ake accelerationde to gra*ity as 1" ms/$)

a) %t the height of -" m0 the metal ball has (!ra+itational potential ener!ykinetic energy)

b) 2st before the metal ball hits the grond0 the ma'imm energy that it has is (gra*itationalpotential energy$inetic ener!y).

c) #alclatei) he energy of the metal ball at the height of-" m.

,!ra+itational = m!h

= &2) ) &40)

= -00

ii) the kinetic energy of the metal before it hits the grond.

,$inetic = ,!ra+itational= -00

d) What is the principle sed in c ii)3

The principle of conser+ation of ener!y

. % motor lifting a weight 1 kg to a height of -." m in - s. he inpt energy spply to the motor in onesecond is $" 2. #alclate

a) power of the motor

o/er = /or$ donetime ta$en

= m!h t

= ) )&4.0) 4

= #0 /att

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4etal ball

-" meter

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b) the efficiency of the motor

,fficiency = seful ener!y output x #001 ,ner!y input

= #0 x #00120

= (0 1

4/)5 A22RECIATING T6E IM2ORTANCE OF MA7IMISING T6E EFFICIENCY OF DE8ICES

1.

force 2.10-2.12

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